Question

Center and Radius of a Sphere Show that the equation represents a sphere, and find its center and radius.

$x^2+y^2+z^2=14y-6z$.

Short answer

The given equation $x^2+y^2+z^2=14y-6z$ is the equation of sphere with center $(0,7,-3)$ and radius $\sqrt{58}$

Step-by-step solution

Step 1. Given information.

Given: $x^2+y^2+z^2=14y-6z$

Step 2. We have to show that the equation represents a sphere, and find its center and radius.

We will complete the squares in x,y and z terms to rewrite the given equation in the form of an equation of sphere.

The given equation is

$\begin{array}{c}{x}^2+y^2+z^2=14y-6z\\ x^2+y^2+z^2-14y+6z=0\\ x^2+\left(y^2-14y+49\right)+\left(z^2+6z+9\right)=0+49+9\\ x^2+{(y-7)}^2+{(z+3)}^2=58\end{array}$

add samenumbers both side.
Comparing this with standard equation of sphere, we can see that the center is $(0,7,-3)$and the radius is $\sqrt{58}$